A cylindrical wire P has resistance 10 Omega | vedclass

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(B) The resistance of a wire is given by the formula $R = \rho \frac{l}{A}$,where $\rho$ is the resistivity,$l$ is the length,and $A$ is the cross-sec

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$20$

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(B) The resistance of a wire is given by the formula $R = ho \frac{l}{A}$,where $ho$ is the resistivity,$l$ is the length,and $A$ is the cross-sectional area. Since $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$,we can write $R = \frac{4 ho l}{\pi d^2}$.For wire $P$: $R_P = 10 \ \Omega$,length $= l$,diameter $= d$.For wire $Q$: length $l_Q = l/2$,diameter $d_Q = d/2$.Since the material is the same,$ho_P = ho_Q = ho$.Calculating the ratio: $\frac{R_Q}{R_P} = \frac{l_Q}{l_P} 	imes \left(\frac{d_P}{d_Q}ight)^2 = \left(\frac{l/2}{l}ight) 	imes \left(\frac{d}{d/2}ight)^2 = \frac{1}{2} 	imes (2)^2 = \frac{1}{2} 	imes 4 = 2$.Therefore,$R_Q = 2 	imes R_P = 2 	imes 10 \ \Omega = 20 \ \Omega$.

$A$ cylindrical wire $P$ has resistance $10 \ \Omega$. $A$ second wire $Q$ has length and diameter half that of $P$. If the material of both the wires is same,then the resistance of wire $Q$ is (in $\Omega$)

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